10
$\begingroup$

Given a finite set of events $\{A_i\}$ which are mutually independent, i.e., for every subset $\{A_n\}$, $$\mathrm{P}\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathrm{P}(A_i).$$

show that the set $\{A_i^c\}$, that is the set of complements of the original events, is also mutually independent.

I can prove this, but my proof relies on the Inclusion-Exclusion principle (as does the proof given in this question). I'm hoping there is a more concise proof.

Can this statement be proved without the use of the Inclusion-Exclusion principle?

$\endgroup$
10
$\begingroup$

Hint: prove that the set of events stays independent if you replace one of them by its complement, i.e. that given your conditions the set $\{A_1^c, A_2, \ldots, A_n\}$ is independent. Then use this $n$ times to replace all of $A_i$ by their complements one by one.

Update. Hint 2: to avoid clutter, let me show you what I mean on the example of two events, $B$ and $C$. Suppose $B$ and $C$ are independent, i.e. $P(B \cap C) = P(B) \cdot P(C)$. We want to show that $B^c$ and $C$ are independent. Indeed: $$ \begin{align*} P(B^c \cap C) &= P(C \setminus (B \cap C)) \\ &= P(C) - P(B \cap C) \\ &= P(C) - P(B) \cdot P(C) \\ &= P(C) \cdot (1-P(B)) \\ &= P(B^c) \cdot P(C). \end{align*} $$ I didn't use inclusion-exclusion here. And this approach scales, i.e. it works the same if you consider more than $2$ variables.

$\endgroup$
  • $\begingroup$ Thanks for the hint, I hadn't thought of doing it this way. But unfortunately I'm still stuck - I can't see how to show $\{A_1^c, A_2, \ldots, A_n\}$ is independent without using the I-E principle. $\endgroup$ – Doria Mar 17 '15 at 0:57
  • $\begingroup$ Great hint Dan! Love this version of the proof. $\endgroup$ – Mehness Apr 26 '18 at 20:44
  • $\begingroup$ Can this argument be used for a contable family of independent events? $\endgroup$ – RLC Feb 3 at 16:18
1
$\begingroup$

Let $[n] = \{1, \cdots, n\}$ and $p(X_1, \cdots, X_n) = \sum_{I \subseteq [n] } a_{I} \prod_{i \in I} X_i$. That is, $p$ is a polynomial in $X_1, \cdots, X_n$ consisting of square-free monomials. If $\{A_1, \cdots, A_n\}$ are mutually independent, then

\begin{align*} \Bbb{E}[p(\mathbf{1}_{A_1}, \cdots, \mathbf{1}_{A_n})] &= \sum_{I \subseteq [n] } a_{I} \Bbb{E}\Big( \prod_{i \in I} \mathbf{1}_{A_i} \Big) \\ &= \sum_{I \subseteq [n] } a_{I} \Bbb{P}\Big( \bigcap_{i \in I} {A_i} \Big) \\ &= \sum_{I \subseteq [n] } a_{I} \prod_{i \in I} \Bbb{P} (A_i) \\ &= p(\Bbb{P}(A_1), \cdots, \Bbb{P}(A_n)), \end{align*}

Then the conclusion follows for the choice $p(X_1, \cdots, X_n) = (1-X_1)\cdots(1-X_n)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.